Question

Problem No-3: The coordinates are shown in units of inches. Assume plane stress conditions. Let E 45x1 displacements have been determined to be u 1-0, v1 0.0025 in., u2 = 0.0012 in, v2 0.0001 in, u3 0, and v3 0.0025 in. Determine: (a) the stiffness matrix for the element shown [k] (b) the element stresses: ??, dy, and ? y and the principal stresses 06 psi, v = 0.25, and thickness t 1 in. The element nodal (15p) (20p) (0. 0) (2. 0)

Answer 1

Given,

E = 45*10⁶ psi

$\nu$ = 0.25

t = 1 in

$k^{e}=t_{e}A_{e}B^{T}DB$

$A_{e}=\frac{1}{2}det\left [ J \right ]=\frac{1}{2}det\begin{bmatrix} x_{13} &y_{13} \\ x_{23}& y_{23} \end{bmatrix}$

$A_{e}=\frac{1}{2}det\begin{bmatrix} 0 &-1\\ 2&-1 \end{bmatrix}$

$\left [ B \right ]=\frac{1}{2A_{e}}\begin{bmatrix} y_{23} & 0 & y_{31} &0 & y_{12} &0 \\ 0& x_{32} & 0 & x_{13} &0 &x_{21} \\ x_{32} &y_{23} & x_{13} & y_{31} & x_{21} & y_{12} \end{bmatrix}$

$\left [ B \right ]=\frac{1}{2*1}\begin{bmatrix} -1 & 0 & 1 &0 & 0 &0 \\ 0& -2 & 0 & 0 &0 &2 \\ -2 &-1 & 0 & 1 & 2 & 0 \end{bmatrix}$

For Plane Stress system

$\left [ D \right ]=\frac{E}{1-\nu ^{2}}\begin{bmatrix} 1 &\nu &0 \\ \nu & 1 & 0\\ 0 & 0 & \frac{1-\nu }{2} \end{bmatrix}$

$\left [ D \right ]=\frac{45*10^{6}}{1-0.25 ^{2}}\begin{bmatrix} 1 &0.25 &0 \\ 0.25 & 1 & 0\\ 0 & 0 & \frac{1-0.25 }{2} \end{bmatrix}$

$\left [ D \right ]=48*10^{6}\begin{bmatrix} 1 &0.25 &0 \\ 0.25 & 1 & 0\\ 0 & 0 &0.375 \end{bmatrix}$

Now

$\left [ k \right ]=1*1*\frac{1}{2}\begin{bmatrix} -1 & 0 &-2 \\ 0& -2 &-1 \\ 1& 0& 0\\ 0& 0 & 1\\ 0& 0 & 2\\ 0&2 &0 \end{bmatrix}*48*10^{6}\begin{bmatrix} 1 & 0.25 &0 \\ 0.25& 1 & 0\\ 0 & 0 & 0.375 \end{bmatrix}*\frac{1}{2}\begin{bmatrix} -1 &0 & 1 & 0 & 0 & 0\\ 0& -2 & 0 & 0 & 0 & 2\\ -2& -1 & 0 &1 & 2& 0 \end{bmatrix}$

$\left [ k \right ]=12*10^{6}\begin{bmatrix} -1 & -0.25 & -0.75\\ -0.5& -2 & -0.375\\ 1& 0.25 &0 \\ 0&0 & 0.375\\ 0& 0 & 0.75\\ 0.5& 2 & 0 \end{bmatrix} \begin{bmatrix} -1 &0 & 1 & 0 & 0 & 0\\ 0 & -2 & 0 & 0 & 0 & 2\\ -2& -1 & 0 & 1 & 2 & 0 \end{bmatrix}$

${\color{Red} \left [ k \right ]=12*10^{6}\begin{bmatrix} 2.5 & 1.25 &-1 &-0.75 &-1.5 &-0.5 \\ 1.25& 4.375& -0.5 & -0.375 & -0.75 &-4 \\ -1& -0.5 &1 & 0 & 0& 0.5\\ -0.75& -0.375 & 0 &0.375 & 0.75 & 0\\ -1.5& -0.75 & 0 & 0.75 & 1.5 & 0\\ -0.5& -4 & 0.5 &0 &0 & 4 \end{bmatrix}}$

Now

$\varepsilon =\left [ B \right ]\left \{ u^{e} \right \}$

$\varepsilon =\left [ B \right ]\begin{Bmatrix} u_{1}\\ v_{1}\\ u_{2}\\ v_{2}\\ u_{3}\\ v_{3} \end{Bmatrix}$

$\varepsilon =\frac{1}{2}\begin{bmatrix} -1 & 0 & 1 & 0 & 0 & 0\\ 0& -2 & 0 & 0 & 0 &2 \\ -2& -1 & 0 & 1 & 2 & 0 \end{bmatrix} \begin{Bmatrix} 0\\ 0.0025\\ 0.0012\\ 0.0001\\ 0\\ 0.0025 \end{Bmatrix}$

$\varepsilon =\frac{1}{2}\begin{Bmatrix} 0.0012\\ 0\\ -0.0024 \end{Bmatrix}$

$\left \{ \sigma \right \}=\left [ D \right ]\left \{ \varepsilon \right \}$

$\begin{Bmatrix} \sigma _{x}\\ \sigma _{y}\\ \tau _{xy} \end{Bmatrix}=48*10^{6}\begin{bmatrix} 1 &0.25 &0 \\ 0.25& 1 & 0\\ 0&0 &0.375 \end{bmatrix} \frac{1}{2} \begin{Bmatrix} 0.0012\\ 0\\ -0.0024 \end{Bmatrix}$

${\color{Red} \begin{Bmatrix} \sigma _{x}\\ \sigma _{y}\\ \tau _{xy} \end{Bmatrix}=\begin{Bmatrix} 28800\\ 7200\\ -21600 \end{Bmatrix}}$

Principle Stresses are given by

$\sigma =\frac{\sigma _{x}+\sigma _{y}}{2}\pm\sqrt{\left ( \frac{\sigma _{x}-\sigma _{y}}{2} \right )^{2}+\tau_{xy}^{2}}$

$\sigma =\frac{28800+7200}{2}\pm\sqrt{\left ( \frac{28800-7200}{2} \right )^{2}+\left ( -21600 \right )^{2}}$

${\color{Red} \sigma _{1}=42149.534\: psi}$

${\color{Red} \sigma _{2}=-6149.534\: psi}$